Find the value of :- $$2010 - \sum_{k=1}^{2010}\bigg\lceil\frac{2010}{k} - \bigg\lfloor \frac{2010}{k}\bigg \rfloor\bigg\rceil $$
What I Tried: Quite stuck at the problem. I need to find some short cut to evaluate the summation but not been able to do so till now.
I know that the factors of $2010$ are $1,2,3,5,6,10,15,30,67,134,201,335,402,670,1005,2010$ . So for all these values in $k$ , the sum of the terms will be $0$ . So we can actually cancel these terms out from the list of $\{1,2,3, ... ,2010\}$ .
The problem is, what about the other numbers? For example, $\frac{2010}{4} = 402.5$ , which can be actually considered $402$ like $5$, but for $\frac{2010}{9} = 223.33$ , it will be considered $223$ , which is actually one number not like $10$ though.
So can anyone help me?
Note that if $k$ is not a divisor of 2010, then
$\bigg\lceil\frac{2010}{k} - \bigg\lfloor \frac{2010}{k}\bigg \rfloor\bigg\rceil=1$
and if $k$ is a divisor, $\bigg\lceil\frac{2010}{k} - \bigg\lfloor \frac{2010}{k}\bigg \rfloor\bigg\rceil=0$
(That's actually because $\frac{2010}{k} - \bigg\lfloor \frac{2010}{k}\bigg \rfloor=${$\frac{2010}{k}$} ({n} is the fractional part of n) and {$\frac{2010}{k}$} is between $0,1$ and is equal to $0$ whenever $\frac{2010}{k}$ is an integer and is equal to $1$ otherwise. So actually $\sum_{k=1}^{2010}\bigg\lceil\frac{2010}{k} - \bigg\lfloor \frac{2010}{k}\bigg \rfloor\bigg\rceil=2010-$(number of positive divisors of 2010)
So the final value you're asking for is equal to the number of positive divisors of 2010 which is 16.